C regex for validating url
This web application is my favorite tool for building, testing, and debugging regular expressions. Writing your own regex expression for this purpose is likely to be an exercise in frustration that will result in eventual (or immediate) disaster when an edge case or minor syntax/grammar inconsistency in the data source causes the expression to fail.I highly recommend that you use it to test out the expressions that we'll cover in this tutorial. Battle-hardened parsers are available for virtually all machine-readable languages, and NLP tools are available for human languages - I strongly recommend that you use one of them instead of attempting to write your own.To demonstrate we'll now quickly go through how to perform this simple regex search on a text file using 16 of the most popular programming languages. Let's go through another simple example - matching any valid year in the 20th or 21st centuries. For instance, hackers can use alternative character encodings to get around naively programmed input blacklist filters.import re import urllib.request import operator # Download wiki page url = "https://en.wikipedia.org/wiki/Diplomatic_history_of_World_War_II" html = urllib.request.urlopen(url).read() # Find all mentioned years in the 20th or 21st century regex = r"\b(? This is another instance where I would strongly recommend using the well-tested libraries and/or services, along with the use of whitelists instead of blacklists, in order to protect your application from malicious inputs.
A regular expression (or regex, or regexp) is a way to describe complex search patterns using sequences of characters. :[\x01-\x08\x0b\x0c\x0e-\x1f\x21-\x5a\x53-\x7f]|\[\x01-\x09\x0b\x0c\x0e-\x7f]) )\]) Yeah, we're not going to walk through that one.There still are lots of regex use cases that we have not covered.For instance, regex can be used in Postgre SQL queries to dynamically search for text patterns within a database.:19|20)\d\b" matches = re.findall(regex, str(html)) # Form a dict of the number of occurrences of each year year_counts = dict((year, matches.count(year)) for year in set(matches)) # Print the dict sorted in descending order for year in sorted(year_counts, key=year_counts.get, reverse=True): print(year, year_counts[year]) . :[\x01-\x08\x0b\x0c\x0e-\x1f\x21\x23-\x5b\x5d-\x7f]|\[\x01-\x09\x0b\x0c\x0e-\x7f])*")@(? Regex matching speeds can range from not-very-fast to extremely slow, depending on how well the expression is written.This allows us to define each part of the pattern as a capture group. This is fine for most use cases, especially if the text being matched is very short (such as an email address form).